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Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for every x

  1. Jun 28, 2008 #1
    Suppose the real valued [tex]g[/tex] is defined on [tex]\mathbb{R}[/tex] and [tex]g'(x) < 0[/tex] for every real [tex]x[/tex]. Prove there's no differentiable [tex]f: R \rightarrow R[/tex] such that [tex]f \circ f = g[/tex].
     
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  3. Jun 28, 2008 #2

    CompuChip

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    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    My suggestion would be to look at (f o f)' (using the chain rule) and try to find some x for which this is necessarily > 0.

    By the way, double posting here was unnecessary and is generally not appreciated here.
     
  4. Jun 28, 2008 #3
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    You want to prove that f is monotone. Then, you want to prove that f isn't monotone.
     
  5. Jun 28, 2008 #4
  6. Jun 28, 2008 #5
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    I don't see it. How do you show that f is bounded?
     
  7. Jun 28, 2008 #6

    matt grime

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    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    It doesn't matter whether f is bounded or not. It is clear that f'(x) and f'(f(x)) have opposite signs for all x. Pick any x. If f(x)=x, then this is a contradiction. If not, then by the intermediate value theorem, f' is zero at some point between x and f(x). That is essentially what the fixed point theorem says.
     
  8. Jun 28, 2008 #7
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    How do you know that f' is continuous?
     
  9. Jun 28, 2008 #8
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    whe only need to know that f is continuous to get f(x) = x, and by assuming that f is differential it is.
     
  10. Jun 28, 2008 #9

    matt grime

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    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    OK, let's assume that the function is once continuously differentiable if that makes you feel better. Off the top of my head I can't think of a function that is differentiable, but where the derivative is discontinuous....
     
  11. Jun 28, 2008 #10
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    The fixed point theorem requires f to be bounded. Otherwise, take f(x) = x + 1

    See wikipidia for a differentiable but not C^1 function.
     
  12. Jun 28, 2008 #11

    matt grime

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    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    If you note, I didn't use the fixed point theorem, just the intermediate value theorem (I've no idea why anyone would invoke the f.p.t. - at best its proof is enlightening for this question). And I suspect that original poster's question was phrased in terms of smooth functions making the exceptional case of a differentiable but not C^1 function important as an aside.
     
  13. Jun 28, 2008 #12
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    my bad didn't see that the fix point theorem needed bounded. matt_grimes proof works, if you assume that f' is continuous.
     
  14. Jun 28, 2008 #13
    [tex]
    A = \{x\in\mathbb{R}\;|\; f'(x)>0\}
    [/tex]

    [tex]
    B = \{x\in\mathbb{R}\;|\; f'(x)<0\}
    [/tex]

    Using continuity of [tex]f[/tex] and checking some preimages, don't you get a contradiction with the fact that the real line is connected?

    edit: Ouch. I wasn't thinking this all the way to the end. Never mind if this is not working.
     
  15. Jun 28, 2008 #14
    Re: Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for eve

    Anyway, my approach was to apply the mean value theorem to show that f is injective. This implies something contradictory about f'.
     
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